Question

(a) You took two 50-g ice cubes from your −15 C kitchen freezer and placed them in 200 g of water in a thermally insulated container. If the water is initially at 25 C, what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used? Hint: Watch out for the different possibilities!

Answer 1

Given mass of ice = 50g=0.1Kg (two ice cubes)

initial temperature of ice=-15 C

mass of water =200 g = 0.2Kg

initial temperature of water =25 C

Heat energy = mass*specific heat* temperature difference

Heat energy = mass*heat of fusion

a)1st teperature of the ice increases from -15 C to 0 C

heat energy =0.10*2.06*15=3.09 J

2nd the ice melts

heat energy =0.2*334=66.8 J

total heat energy =3.09+66.8=69.89 J

69.89=mass* specific of water *change in temprature

69.89=0.2*4.18*change in temprature

change in temprature=83.60

temperature of the hot water =83.60-25=58.6 C

b) 1st teperature of the ice increases from -15 C to 0 C

heat energy =0.050*2.06*15=1.545 J

2nd the ice melts

heat energy =0.2*334=66.8 J

total heat energy =1.545+66.8=68.345 J

68.345=mass* specific of water *change in temprature

68.345=0.2*4.18*change in temprature

change in temprature=81.75

temperature of the hot water =81.75-25=56.75C